I'm trying understand how to compute evolution equation for $g_{\sigma, \eta}$ defined on page $3$ of this article and I'm reading the Lectures on Mean Curvature Flows by Xi-Ping Zhu, which gives some details about the computation, I could understand all the computations, except for the computation of the Laplacian, which is
\begin{align*} \Delta g_{\sigma,\eta} &= \frac{1}{H^{2 - \sigma}} \left[ \Delta |A|^2 - \sigma (1+\eta) H \Delta H + (1 + \eta) \sigma (\sigma - 3) |\nabla H|^2 \right]\\ &- 2 (2 - \sigma) \frac{\nabla_i H}{H} \cdot \nabla_i g_{\sigma,\eta} + \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \Delta H \right] \end{align*}
according to the book (the author use '$\cdot$' to represent the inner product ), but my computation differ from the above by the terms with $(1 + \eta)$. I would like to know how I can obtain the Laplacian correctly.
This is what I did
$\triangle g_{\sigma, \eta} = \nabla_i \nabla_i g_{\sigma,\eta}$
$= \{ \left[ \triangle |A|^2 -2(1+\eta)(|\nabla H|^2 + H\triangle H) \right] H^{2- \sigma} - (\nabla_i |A|^2 - 2(1 + \eta) H \nabla_i H) [ (2 - \sigma) H^{1 - \sigma} \nabla_i H ] \} \frac{1}{(H^{2 - \sigma})^2} - (2 - \sigma) \{ \left( \frac{|A|^2 - (1 + \eta) H^2}{H^{3 - \sigma}} \right) \triangle H \left[ (\nabla_i |A|^2 - (1 + \eta) 2H \nabla_i H) H^{3 - \sigma} - (|A|^2 - (1 + \eta)H^2) (3- \sigma) H^{2 - \sigma} \nabla_i H \right]$
$\frac{\nabla_i H}{(H^{3 - \sigma})^2} \}$
$= \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} - (\sigma^2 - 3\sigma + 4) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2 \frac{(2 - \sigma)}{H^{2 - \sigma}} \left\langle \nabla_i |A|^2, \frac{\nabla_i H}{H} \right\rangle$
$- \frac{(2 - \sigma) |A|^2 H \triangle H}{H^{4 - \sigma}} + \frac{(2 - \sigma)(3 - \sigma) |A|^2 |\nabla H|^2}{H^{4 - \sigma}} (\star)$
Observe that
$-2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle = - \frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 4(2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}$
$\frac{2(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}} - \frac{2(2 - \sigma)^2 (1 + \eta) H^2 |\nabla H|^2}{H^{4 - \sigma}}$
$= -\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} + 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}},$
therefore
$-\frac{2(2 - \sigma)}{H^{2 - \sigma}} \left\langle \frac{\nabla_i H}{H}, \nabla_i |A|^2 \right\rangle = -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle - 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} - 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}}$
Substituting this in $(\star)$,
$\triangle g_{\sigma,\eta} = \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} - (-\sigma^2 + \sigma + 4) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle$
$+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \triangle H \right].$
Thanks in advance!
Computing in normal coordinates, \begin{align*} \Delta g_{\sigma, \eta} &= \nabla_i \nabla_i g_{\sigma,\eta}\\ &= \nabla_i \left[ \frac{(\nabla_i |A|^2 - (1 + \eta) 2 H \nabla_i H) H^{2 - \sigma} - (|A|^2 - (1 + \eta) H^2) (2 - \sigma) H^{1 - \sigma} \nabla_i H}{(H^{2 - \sigma})^2} \right]\\ &= \nabla_i \left[ \frac{(\nabla_i |A|^2 - 2 (1 + \eta) H \nabla_i H}{H^{2 - \sigma}} - (2 - \sigma) \frac{(|A|^2 - (1 + \eta) H^2) \nabla_i H}{H^{3 - \sigma}} \right]\\ &= \left[ (\Delta |A|^2 - 2 (1 + \eta) (H \Delta H + |\nabla H|^2)) H^{2 - \sigma} - (\nabla_i |A|^2 - 2 (1 + \eta) H \nabla_i H) \right.\\ &\left. (2 - \sigma) H^{1 - \sigma} \nabla_i H \right] \frac{1}{(H^{2 - \sigma})^2}\\ &- (2 - \sigma) \left[ (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle + |A|^2 \Delta H - (1 + \eta) (2 H |\nabla H|^2 + H \Delta H)) H^{3 - \sigma} \right.\\ &\left. - (|A|^2 \nabla_i H - (1+ \eta) H^2 \nabla_i H) (3 - \sigma) H^{2 - \sigma} \nabla_i H \right] \frac{1}{(H^{3 - \sigma})^2}\\ &= \left[ (\Delta |A|^2 - 2 (1 + \eta) (H \Delta H + |\nabla H|^2)) H^{2 - \sigma} - (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle - 2 (1 + \eta) H |\nabla H|^2) \right.\\ &\left. (2 - \sigma) H^{1 - \sigma} \right] \frac{1}{(H^{2 - \sigma})^2}\\ &- (2 - \sigma) \left[ (\left\langle \nabla_i |A|^2, \nabla_i H \right\rangle + |A|^2 \Delta H - (1 + \eta) (2 H |\nabla H|^2 + H \Delta H)) H^{3 - \sigma} \right.\\ &\left. - (|A|^2 |\nabla H|^2 - (1+ \eta) H^2 |\nabla H|^2) (3 - \sigma) H^{2 - \sigma} \right] \frac{1}{(H^{3 - \sigma})^2}\\ &= \left[ \Delta |A|^2 - \sigma (1 + \eta) H \Delta H + (- \sigma^2 + \sigma) (1 + \eta) |\nabla H|^2 \right] \frac{1}{(H^{2 - \sigma})^2} + \frac{2 \left\langle \nabla_i |A|^2, \nabla_i H \right\rangle}{H^{3 - \sigma}}\\ &+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (3 - \sigma) - H |A|^2 \Delta H \right] (\star) \end{align*}
Observe that \begin{align*} -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle &= - \frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 4(2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\ &+ \frac{2(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}} - \frac{2(2 - \sigma)^2 (1 + \eta) H^2 |\nabla H|^2}{H^{4 - \sigma}}\\ &= -\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} + 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\ &+ 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}}, \end{align*} i.e., \begin{align*} -\frac{2(2 - \sigma) \left\langle \nabla_i H, \nabla_i |A|^2 \right\rangle}{H^{3 - \sigma}} &= -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle - 2 \sigma (2 - \sigma) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}}\\ &- 2 \frac{(2 - \sigma)^2 |A|^2 |\nabla H|^2}{H^{4 - \sigma}} \end{align*} Substituting this in $(\star)$, \begin{align*} \Delta g_{\sigma,\eta} &= \frac{\triangle |A|^2}{H^{\sigma - 2}} - \frac{\sigma(1 + \eta) H \triangle H}{H^{\sigma - 2}} + \sigma (\sigma - 3) (1 + \eta) \frac{|\nabla H|^2}{H^{2 - \sigma}} -2(2 - \sigma) \left\langle \frac{\nabla_i H}{H}, \nabla_i g_{\sigma, \eta} \right\rangle\\ &+ \frac{(2 - \sigma)}{H^{4 - \sigma}} \left[ |A|^2 |\nabla H|^2 (\sigma - 1) - H |A|^2 \triangle H \right]. \end{align*}